Deducing $3^f = o(3^g)$ from $f = o(g)$

I really need help solving the following question:

Given: $$f(n) = o(g(n))$$

Prove: $$3^{f(n)} = o(3^{g(n)})$$

My attempt:

I know that $$\frac{f(n)}{g(n)} \xrightarrow{} 0$$.

I need to prove that $$f(n) - g(n) \xrightarrow{} -\infty$$ so that $$3^{f(n)-g(n)} \xrightarrow{} 0$$.

How do I prove that?

In general, I'm not sure what properties can I assume about $$f$$ and $$g$$. Are they positive? What else do I know about them? Are we assuming that all these function approach $$\infty$$ or can they be $$\frac{1}{n}$$ etc.?

• You should direct your questions to whoever set you this exercise. – Yuval Filmus Jul 6 '19 at 11:04
• If we assume that $f,g$ take values in the positive integers, then $f = o(g)$ does imply $f - g \to -\infty$. – Yuval Filmus Jul 6 '19 at 12:35
• @YuvalFilmus Isn't it necessary to say that both functions approach $\infty$ in order to say that? And if not, how did you deduce it? I'm self studying this and I'm using material from a friend. – PhysicsPrincess Jul 6 '19 at 16:12
• Is it actually true that all function that take value in the positive integers approach $\infty$ (I can't come up with one that doesn't) – PhysicsPrincess Jul 6 '19 at 16:19
• How about the function $f(n) = 1$? However, if $f = o(g)$ and $f,g$ takes values in the positive integers, then since $f \geq 1$, we must have $g \to \infty$. – Yuval Filmus Jul 6 '19 at 16:21